OFFSET
0,3
COMMENTS
A375495(k) is the number of ways to reach k by the minimum length construction and the present sequence is those k for which A375495(k) = 1.
This sequence is infinite, since k = f(f(f(...(1)))) (A003462) is always a unique minimum construction.
Experimentally, it is observed that many of the integers with unique smallest constructions have one or more g(x) steps in their construction and the position of those g(x) operations does not follow an obvious pattern; meaning that these unique constructions are complex and may be related to the complexity of the Hailstone sequences (A127933).
EXAMPLE
9 is a term since the shortest sequence of f and g to reach 9 (length A375494(9) = 5) is unique g(f(g(f(f(1))))) = 9.
Here are some of the unique, smallest constructions. All other positive integers smaller than 28 do not have unique, smallest constructions.
0 = g(1)
1 = 1
2 = g◦f(1)
4 = f(1)
6 = g◦f◦f(1)
7 = f◦g◦f(1)
9 = g◦f◦g◦f◦f(1)
11 = g◦f◦f◦g◦f(1)
13 = f◦f(1)
17 = g◦f◦g◦f◦f◦g◦f(1)
19 = f◦g◦f◦f(1)
20 = g◦f◦f◦f(1)
22 = f◦f◦g◦f(1)
26 = g◦f◦g◦f◦g◦f◦f◦g◦f(1)
28 = f◦g◦f◦g◦f◦f(1)
PROG
(Python)
from itertools import product
seq = [None for _ in range(200)]
num = [ 0 for _ in range(len(seq))]
for L in range(0, 23):
for P in product((True, False), repeat=L):
x = 1
for upward in P:
x = 3*x+1 if upward else x//2
if x < len(seq):
if num[x] == 0 or L < seq[x]:
seq[x], num[x] = L, 1
elif L == seq[x]:
num[x] += 1
print(', '.join([str(i) for i, x in enumerate(num) if x==1]))
CROSSREFS
KEYWORD
nonn
AUTHOR
Russell Y. Webb, Aug 31 2024
STATUS
approved